An apartment has a living room whose dimensions are 2.5 m 4.8 m 5.9 m. Assume that the air in the room is composed of 79% nitrogen (N2) and 21% oxygen (O2). At a temperature of 27°C and a pressure of 1.01 105 Pa, what is the mass (in grams) of the air?
3 answers
PV = nRT
volume = 2.5 x 4 x 5 = 50
temperature = 22+273.15 = 295.15 k
pressure = 1.01x 10^5 pa
79% of N2 = 0.79
21% of O2 = 0.21
pv = nrt
1.01 x 10^5 x 50 = n x 8.31 x 295.15
n= 2059
n = mass / mass over mole
mass= 2059 x (0.79 x 14*2+0.21 *32)
mass = 5.9*10^4
temperature = 22+273.15 = 295.15 k
pressure = 1.01x 10^5 pa
79% of N2 = 0.79
21% of O2 = 0.21
pv = nrt
1.01 x 10^5 x 50 = n x 8.31 x 295.15
n= 2059
n = mass / mass over mole
mass= 2059 x (0.79 x 14*2+0.21 *32)
mass = 5.9*10^4
oh just change the temperature and the pressure in my equation. I got the wrong question but overall method is the same. Good luck!
1 answer